A fast pitch softball player does a \"windmill\" pitch, moving her hand through

Question

A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 70{ m mph} . The 0.18{ m kg} ball is 54{ m cm} from the pivot point at her shoulder.

A) Just before the ball leaves her hand, what is its centripetal acceleration?
B) At the lowest point of the circle the ball has reached its maximum speed. What is the magnitude of the force her hand exerts on the ball at this point?
C) At the lowest point of the circle the ball has reached its maximum speed. What is the direction of the force her hand exerts on the ball at this point?

Explanation / Answer

70 mph = 31.2928 m/s a) ac = V²/r = 31.2928²/.54 = 1813.4 m/s² b) at = (dv)²/(2*dx) = 31.2928²/(2*2p*.54) = 144.379 m/s²; atot = v(ac²+at²) = v(1813.4²+144.379²) = 1819.138 m/s² F = m*atot = .18*1819.138 = 327.444 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.